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Bài làm :
\(a\text{)}3x^2+4x=0\Leftrightarrow x\left(3x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\3x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{4}{3}\end{cases}}\)
\(b\text{)}25x^2-0,64=0\Leftrightarrow\left(5x-0,8\right)\left(5x+0,8\right)=0\Leftrightarrow\orbr{\begin{cases}5x-0,8=0\\5x+0,8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0,16\\-0,16\end{cases}}\)
\(c\text{)}x^4-16x^2=0\Leftrightarrow\left(x^2-4x\right)\left(x^2+4x\right)=0\Leftrightarrow\orbr{\begin{cases}x^2-4x=0\\x^2+4x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x-4\right)=0\\x\left(x+4\right)=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
\(d\text{)}x^2+x=6\Leftrightarrow x^2+x-6=0\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
Bài làm :
\(a)3x^2+4x=0\)
\(\Rightarrow x\left(3x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\3x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-4}{3}\end{cases}}\)
Vậy x = 0 hoặc \(x=\frac{-4}{3}\) .
\(b)25x^2-0,64=0\)
\(\Rightarrow\left(5x\right)^2=\frac{16}{25}\)
\(\Rightarrow\left(5x\right)^2=\left(\frac{4}{5}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}5x=\frac{4}{5}\\5x=\frac{-4}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{4}{25}\\x=\frac{-4}{25}\end{cases}}\)
Vậy \(x=\frac{4}{25}\) hoặc \(x=\frac{-4}{25}\) .
\(c)x^4-16x^2=0\)
\(\Rightarrow x^2\left(x^2-16\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2=0\\x^2-16=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=4^2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
Vậy x = 0 hoặc \(x=\pm4\) .
đặt x^2-7x=y=> \(y\ge-\frac{49}{4}\) (*)
\(A=y\left(y+12\right)=y^2+12y=\left(y+6\right)^2-36\ge-36\)
đẳng thức khi y=-6 thủa mãn đk (*)
Vậy: GTNN của A=-36 khí y=-6 =>\(\left[\begin{matrix}x=1\\x=6\end{matrix}\right.\)
a) x3 + 3x2 + 3x + 1 = 64
=> (x + 1)3 = 64
=> (x + 1)3 = 43
=> x + 1 = 4 => x = 3
b) x3 + 6x2 + 9x = 4x
=> x3 + 6x2 + 9x - 4x = 0
=> x3 + 6x2 + 5x = 0
=> x3 + 5x2 + x2 + 5x = 0
=> x2(x + 5) + x(x + 5) = 0
=> (x + 5)(x2 + x) = 0
=> (x + 5)x(x + 1) = 0
=> \(\hept{\begin{cases}x=-5\\x=0\\x=-1\end{cases}}\)
c) 4(x - 2)2 = (x + 2)2
=> 4(x2 - 4x + 4) = x2 + 4x + 4
=> 4x2 - 16x + 16 = x2 + 4x + 4
=> 4x2 - 16x + 16 - x2 - 4x - 4 = 0
=> 3x2 - 20x + 12 = 0
=> 3x2 - 18x - 2x + 12 = 0
=> 3x(x - 6) - 2(x - 6) = 0
=> (x - 6)(3x - 2) = 0
=> \(\orbr{\begin{cases}x=6\\x=\frac{2}{3}\end{cases}}\)
d) x4 - 16x2 = 0
=> x2(x2 - 16) = 0
=> \(\orbr{\begin{cases}x^2=0\\x^2=16\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
e) x4 - 4x3 + x2 - 4x = 0
=> x4 + x2 - 4x3 - 4x = 0
=> x2(x2 + 1) - 4x(x2 + 1) = 0
=> (x2 - 4x)(x2 + 1) = 0
=> x(x - 4)(x2 + 1) = 0
=> \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)(vì x2 + 1 \(\ge\)1 > 0 \(\forall\)x)
f) x3 + x = 0 => x(x2 + 1) = 0 => x = 0 (vì x2 + 1 \(\ge1>0\forall\)x)
1.
a) \(2x\left(x-4\right)+\left(x-1\right)\left(x+2\right)=2x^2-8x+x^2+x-2=x^2-7x-2\)
b) \(\left(x-3\right)^2-\left(x-2\right)\left(x^2+2x+4\right)=x^2-6x+9-x^3+8=-x^3+x^2-6x+17\)
2.
a) \(x^2y+xy^2-3x+3y=xy\left(x+y\right)-3\left(x-y\right)=???\)
b) \(x^3+2x^2y+xy^2-16x=x\left(x^2+2xy+y^2-16\right)=x\left[\left(x+y\right)^2-16\right]=\)làm tiếp chắc dễ
3.
\(\frac{x^4?2x^3+4x^2+2x+3}{x^2+1}\) Giữa x^4 và 2x^3 (vị trí dấu ? là dấu + hay -)
4) \(A=x^2-3x+4=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\)
\(A\ge\frac{7}{4}\)
Vậy GTNN của A là 7/4
a)\(a^4+a^2+1=\left(a^2\right)^2+2a^2.1+1^2-a^2=\left(a^2+1\right)^2-a^2=\left(a^2+1+a\right)\left(a^2+1-a\right)\)
b)\(a^4+a^2-2=a^4-a^2+2a^2-2=a^2\left(a^2-1\right)+2\left(a^2-1\right)=\left(a^2+2\right)\left(a^2-1\right)\)
c)\(x^4+4x^2-5=x^4-x^2+5x^2-5=x^2\left(x^2-1\right)+5\left(x^2-1\right)=\left(x^2+5\right)\left(x+1\right)\left(x-1\right)\)
d)\(\left(x+2\right)\left(x^2-2x-6\right)=x^3-2x^2-6x+2x^2-4x-12=x^3-10x-12\)
\(\Rightarrow x^3-10x-12=\left(x+2\right)\left(x^2-2x-6\right)\)
e)\(6x^3-17x^2+14x-3\)
Ta có: \(\left(ax^2+bx+c\right)\left(dx+e\right)\)
\(=adx^3+aex^2+bdx^2+bex+cdx+ce\)
\(=adx^3+\left(ae+bd\right)x^2+\left(be+cd\right)x+ce\)
Do đó:\(\left\{{}\begin{matrix}ad=6\\ae+bd=-17\\be+cd=14\\ce=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=3;b=-4\\c=1;d=2\\e=-3\end{matrix}\right.\)
Suy ra: \(6x^3-17x^2+14x-3=\left(3x^2-4x+1\right)\left(2x-3\right)\)
h)\(x^4-34x^2+225=x^4-15x^2-15x^2+225-4x^2=x^2\left(x^2-15\right)-15\left(x^2-15\right)-\left(2x\right)^2=\left(x^2-15\right)^2-\left(2x\right)^2=\left(x^2+2x-15\right)\left(x^2-2x-15\right)=\left(x^2-3x+5x-15\right)\left(x^2+5x-3x-15\right)=\left[\left(x-3\right)\left(x+5\right)\right]^2\)